I'm actually using the same idea I used in China TST 2023 P16 [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.617809341722198, xmax = 9.047803017513637, ymin = -2.881439029044769, ymax = 9.287510408972135; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw(circle((-5,3.5), 3.5), linewidth(2)); draw(circle((-0.937805683371705,5.821326952425198), 1.1786730475748013), linewidth(2)); draw(circle((1.0312230509779012,2.598260820760513), 2.598260820760513), linewidth(2)); draw((5.49613768265097,0)--(-10.792093230726477,0), linewidth(2)); draw((5.49613768265097,0)--(-10.792093230726477,0), linewidth(2)); draw((5.49613768265097,7)--(-10.792093230726477,7), linewidth(2)); draw((-0.9378056420763958,7)--(-5,0), linewidth(1.6) + linetype("2 2") + blue); draw((1.0312230509779012,0)--(-5,7), linewidth(1.6) + linetype("4 4") + blue); draw((-5,7)--(-5,0), linewidth(2)); draw((-0.9378056420763958,7)--(1.0312230509779012,0), linewidth(1.6) + linetype("4 4") + ccqqqq); /* dots and labels */ dot((-5,7),dotstyle); label("$F$", (-4.921286869914071,7.181346083161517), NE * labelscalefactor); dot((-5,0),dotstyle); label("$G$", (-4.921286869914071,0.1787997349535647), NE * labelscalefactor); dot((-5,3.5),linewidth(4pt) + dotstyle); dot((-1.9611725453718565,5.236527483513647),dotstyle); label("$C$", (-1.8970509149039485,5.417208442738948), NE * labelscalefactor); dot((-0.937805683371705,5.821326952425198),linewidth(4pt) + dotstyle); dot((1.0312230509779012,2.598260820760513),linewidth(4pt) + dotstyle); dot((-0.9378056420763958,7),linewidth(4pt) + dotstyle); label("$A$", (-0.8709708587397998,7.145343274173301), NE * labelscalefactor); dot((1.0312230509779012,0),linewidth(4pt) + dotstyle); label("$B$", (1.1091836356120661,0.142796925965349), NE * labelscalefactor); dot((-1.5384752638720163,2.982461111437899),linewidth(4pt) + dotstyle); label("$D$", (-1.4650172070453595,3.1310300719872517), NE * labelscalefactor); dot((-0.32332812565340757,4.815500185959009),linewidth(4pt) + dotstyle); label("$E$", (-0.2589231059401322,4.967173330386252), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] First, Let $\omega$ touching $\ell_1$ and $\ell_2$ at $F$ and $G$ respectively. It is easy to see that $FG$ is diameter of $\omega$ Claim 1 : $\overline{G,C,A}$ (similarly $\overline{F,D,B}$) Proof. Since $\angle FCG$ = $90^\circ$ then it is suffices to show that $\angle FCA$ = $90^\circ$ Let common tangents of $\omega$ and $\omega_1$ intersects $\ell_1$ at $K$ It is known that $KF$ = $KC$ = $KA$. Thus $\angle FCA$ = $90^\circ$ as we need. $\square$ Claim 2 : $\overline{A,E,B}$ Proof. Let common tangents of $\omega_1$ and $\omega_2$ intersects $\ell_1$ and $\ell_2$ at $H$ and $I$ respectively then, $$\angle AEH = 90^\circ - \frac{\angle AHE}{2} = 90^\circ - \frac{\angle BIE}{2} = \angle BEM$$Thus the claim is proven. $\square$ Claim 3 : $A,E,D,F$ and $C,E,B,G$ are concyclic Proof. Just angle chasing, $$\angle AFD = \angle DBG = \angle DEB$$implies $A,E,D,F$ are concyclic and, $$\angle AEC = \angle FCA = \angle AGB = \angle CGB$$implies $C,E,B,G$ are concyclic. $\square$ By above claim, we obtained that $Pow_B$($\omega$) = $BD \times BF$ = $BE \times BA$ = $Pow_B$($\omega_1$) then $B$ lies on radical axis of $\omega$ and $\omega_1$ Thus $BC$ is radical axis of $\omega$ and $\omega_1$. similarly $AD$ is radical axis of $\omega$ and $\omega_2$ Implied radical axis theorem on $\omega$ , $\omega_1$ and $\omega_2$ then we are done. $\blacksquare$

I will prove that $\overline{BC}$ passes through the circumcenter of $\triangle CDE$, which is enough. Invert at $C$; because the circumcenter and orthocenter are isogonal conjugates, the problem becomes the following (totally changing object labels, but preserving point labels). Restated wrote: Let $\ell_1$ and $\ell_2$ be two parallel lines. Let $\omega_1$ be a circle tangent to $\ell_1$ and $\ell_2$ respectively at $D$ and $E$. Let $\omega$ be a circle tangent to $\ell_1$ and $\omega_1$ at $B$. Let $\omega_2$ be a circle tangent to $\ell_1$ and $\ell_2$, as well as to $\omega$ at $C$. Prove that $\overline{BC} \perp \overline{DE}$. This is equivalent to $\overline{BC} \parallel \ell_1$, which is clear by symmetry.